Computing the probability of voting of the member of Parliament

[Dhamnekar Win,odd]

Parliament contains a proportion p of Labour members, who are incapable of changing their minds about anything, and a proportion 1 − p of Conservative members who change their minds completely at random (with probability r) between successive votes on the same issue. A randomly chosen member is noticed to have voted twice in succession in the same way. What is the probability that this member will vote in the same way next time?

 

My answer:

Let's solve this problem step by step. 

First, let's find the probability that the member is a Labour member given that they have voted twice in succession in the same way. We can use Bayes' theorem for this. Let L be the event that the member is a Labour member and V be the event that the member has voted twice in succession in the same way. Then, we have:

P(L|V) = P(V|L) * P(L) / P(V)

Since a Labour member is incapable of changing their mind, P(V|L) = 1. The prior probability of a member being a Labour member is given as p, so P(L) = p. To find P(V), we can use the law of total probability:

P(V) = P(V|L) * P(L) + P(V|L^c) * P(L^c)

where L^c is the event that the member is not a Labour member (i.e., they are a Conservative member). As mentioned earlier, P(V|L) = 1 and P(L) = p. Since a Conservative member changes their mind completely at random with probability r between successive votes on the same issue, the probability that they vote in the same way twice in succession is (1-r), so P(V|L^c) = (1-r). The prior probability of a member being a Conservative member is given as (1-p), so P(L^c) = (1-p). Plugging these values into the equation above, we get:

P(V) = 1 * p + (1-r) * (1-p)

Now we can plug this value into our equation for P(L|V):

P(L|V) = 1 * p / [p + (1-r)*(1-p)]

Now that we have found the probability that the member is a Labour member given that they have voted twice in succession in the same way, we can find the probability that they will vote in the same way next time. Since a Labour member is incapable of changing their mind and a Conservative member changes their mind completely at random with probability r between successive votes on the same issue, this probability is:

P(vote in same way next time | V) = P(L|V) * 1 + [1 - P(L|V)] * (1-r)

Plugging in our value for P(L|V), we get:

P(vote in same way next time | V) = [p / [p + (1-r)*(1-p)]] * 1 + [1 - p / [p + (1-r)*(1-p)]] * (1-r)

This simplifies to:

P(vote in same way next time | V) = 1 - r(1-p)/(p + (1-r)(1-p))

This is our final answer.

Is this answer correct?

Note: credit for this answer goes to Microsoft Artificial Intelligence  powered chat.

Edited April 23, 2023 by Dhamnekar Win,odd

[Genady]

14 hours ago, Dhamnekar Win,odd said:

P(vote in same way next time | V) = [p / [p + (1-r)*(1-p)]] * 1 + [1 - p / [p + (1-r)*(1-p)]] * (1-r)

This simplifies to:

P(vote in same way next time | V) = 1 - r(1-p)/(p + (1-r)(1-p))

I don't think it is correct. Check the algebra.

[Dhamnekar Win,odd]

7 hours ago, Genady said:

I don't think it is correct. Check the algebra.

Yes. You are correct. The final answer is [math] 1-\displaystyle\frac{r(1-p)(1-r)}{(p + (1-r)(1-p))}[/math]

Edited April 24, 2023 by Dhamnekar Win,odd

[John Cuthber]

22 hours ago, Dhamnekar Win,odd said:

Parliament contains a proportion p of Labour members, who are incapable of changing their minds about anything

No, it doesn't.
If you want to discuss statistics that's fine.
But starting with a false statement is just going to distract people.

This
 

22 hours ago, Dhamnekar Win,odd said:

proportion 1 − p of Conservative members who change their minds completely at random

isn't any better.

[Genady]

7 minutes ago, Dhamnekar Win,odd said:

Yes. You are correct. The final answer is 1−r(1−p)(1−r)(p+(1−r)(1−p))

Still incorrect.

[Dhamnekar Win,odd]

22 minutes ago, John Cuthber said:

No, it doesn't.
If you want to discuss statistics that's fine.
But starting with a false statement is just going to distract people.

This
 

isn't any better.

It is given in the question ab initio. 

20 minutes ago, Genady said:

Still incorrect.

 It is the division in the second term, not multiplication.

[Genady]

56 minutes ago, Dhamnekar Win,odd said:

 It is the division in the second term, not multiplication.

I know.

I mean, the formula as it appears here

https://www.scienceforums.net/topic/131457-computing-the-probability-of-voting-of-the-member-of-parliament/?do=findComment&comment=1238311

is incorrect.

Edited April 24, 2023 by Genady

[Dhamnekar Win,odd]

9 hours ago, Genady said:

I don't think it is correct. Check the algebra.

If we solve,

P(vote in same way next time | V) = [p / [p + (1-r)*(1-p)]] * 1 + [1 - p / [p + (1-r)*(1-p)]] * (1-r)

This simplifies to:

So,final answer ishttps://www.scienceforums.net/topic/131457-computing-the-probability-of-voting-of-the-member-of-parliament/?do=findComment&comment=1238311

 

 

[Genady]

2 hours ago, Dhamnekar Win,odd said:

If we solve,

P(vote in same way next time | V) = [p / [p + (1-r)*(1-p)]] * 1 + [1 - p / [p + (1-r)*(1-p)]] * (1-r)

This simplifies to:

No, these two expressions above certainly do not equal each other. I can prove it.

Moreover, the second expression cannot be a probability at all.

[Dhamnekar Win,odd]

1 hour ago, Genady said:

No, these two expressions above certainly do not equal each other. I can prove it.

Moreover, the second expression cannot be a probability at all.

Read here  the whole computation:

[math]\displaystyle\frac{p}{p+(1-r)(1-p)} \cdot 1 + \left[1-\displaystyle\frac{p}{p+(1-r)(1-p)}\right]\cdot (1-r)= -\displaystyle\frac{p\cdot(r^2-2\cdot r)-(r^2-2r +1)}{r \cdot p -(r-1)}= 1-\displaystyle\frac{r\cdot (1-r) \cdot (1-p)}{p + (1-r)\cdot (1-p)}[/math] 

 

 

Edited April 24, 2023 by Dhamnekar Win,odd

[Genady]

Sorry, you are right. I didn't notice the minus sign in this image:

4 hours ago, Dhamnekar Win,odd said:

My apologies.

The final expression in the OP was incorrect, but you have corrected it.